Thermovid 8 01

Thermovid 8 01

Statistical Molecular Thermodynamics Christopher J. Cramer Video 8.1 Helmholtz Free Energy Spontaneity Redux is the condition for spontaneity only for an dS ≥ 0 isolated system (constant U and V) If the system is not isolated then we must consider the entropy of both the system and the surroundings to evaluate the total change ΔStotal, which is not very convenient. Consider a system at constant T and V. The system is not isolated since heat can flow to maintain constant T. As a result dS ≥ 0 for the system alone is not the criterion for a spontaneous process. So, what is the criterion for a spontaneous process at constant T and V? Spontaneity at Constant T and V From the second law: At constant V: δq dS ≥ → δq ≤ TdS δw = −Pext dV = 0 T The first law: dU = δq +δw irreversible dU ≤ TdS (constant V) reversible Note that for an isolated system dU = 0 and we recover dS ≥ 0 The Helmholtz Free Energy dU ≤ TdS dU −TdS ≤ 0 (constant V) irreversible d(U − TS) ≤ 0 (constant T and V) Hermann von Helmholtz reversible Define a new state function, the Helmholtz free energy, A: A = U −TS € spontaneous A (constant T and V) dA ≤ 0 Equilibrium Amin at equilibrium time Helmholtz “Whoever, in the pursuit of science, seeks after immediate practical utility, may generally rest assured that he will seek in vain.” (inventor of the opthalmoscope) Physician and professor of physiology and psychology who did fundamental work on the transmission of nerve signals, Hermann von Helmholtz opthalmology, auditory perception, oh, and thermodynamics… Academic appointments at Königsberg, Bonn, Heidelberg, and, ultimately, as Professor of Physics at Berlin, where he was a senior colleague of Max Planck The Helmholtz Free Energy A = U −TS (definition) ΔA = ΔU −TΔS (isothermal change) spontaneous dA ≤ 0 ΔA = ΔU −TΔS ≤ 0 (constant T and V) at equilibrium For a process to take place spontaneously there is a compromise between lowering the energy and increasing the entropy. Entropy becomes more important at high T. A process where ΔA is positive will not take place spontaneously at constant T and V. In order for such a process to take place you must add something such as work. Isothermal Work from a Spontaneous Process ΔA = ΔU −TΔS A is a state function, so to evaluate it we can take a reversible q path from one state to another ΔS = rev → TΔS = q T rev ΔA = ΔU − qrev (isothermal, reversible) ΔA = wrev For ΔA < 0, the process is spontaneous and wrev is the maximum (free) work (Arbeit) that can be extracted (–wrev > –wirr). For ΔA > 0, the process is not spontaneous and wrev is the minimum work that must be done to drive the process. .

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